Bound on the index of an abelian subgroup in discrete subgroup of the euclidean group? I asked the following question on math.SE a couple of days ago. Dietrich Burde gave an answer for the case that the subgroup is not only discrete but also acts cocompactly.
What about the general case? Are there any upper bounds known?

$\DeclareMathOperator{\isom}{Isom}$A discrete subgroup of the group of isometries in euclidean space is almost abelian.
By this I mean that for each $n$ there exists $m$ such that for any discrete subgroup $\Gamma$ of $\isom(\mathbb{R}^n)$ we can find an abelian subgroup $\Gamma' \leq \Gamma$ such that $$[\Gamma : \Gamma'] \leq m,$$ so the index of the abelian subgroup in $\Gamma$ is bounded by $m$.
What is the best known bound on $m$?
 A: Every discrete group of Euclidean isometries acts properly discontinuously and cocompactly on some subspace of $\mathbb{R}^n$ (a result of Bieberbach). If we are in the crystallographic case, then the index is bounded by the maximal order of a finite subgroup of $GL(n,\mathbb{Z})$. Here we have several results in the literature, e.g., the result of Friedland in $1997$, that
$$m\le 2^nn!
$$ 
for $n\ge n_0$.
There are exceptions for small $n$, see the comment of Geoff Robinson.
There are conditions given in Friedland's paper "The maximal orders of finite subgroups of $GL(n,\mathbb{Q})$", when equality is attained.
Rockmore in $1995$ proved the following: for every $\epsilon>0$ there exists a constant $c(\epsilon)$ such that $m\le c(\epsilon)(n!)^{1+\epsilon}$.
A: Since ${\rm Isom}(\mathbb R^n) = \mathbb R^n \rtimes O(n) \subset GL(n+1,\mathbb C)$, the statement can be reduced to subgroups of $GL(n+1,\mathbb C)$. It was proved in 
Boris Weisfeiler, Post-classification version of Jordan's theorem on finite linear groups, Proc. Natl. Acad. Sci. USA. Aug 1984; 81(16): 5278–5279.
that any finite subgroup of $GL(n,\mathbb C)$ has an abelian normal subgroup of index at most $(n+1)! n^{a log(n) + b}$. I would expect that the same result holds for virtually abelian subgroups, so that you would also obtain a similar bound in your situation.
Looking at
Eli Aljadeff and Jack Sonn, Bounds on orders of finite subgroups of ${\rm PGL}_n(K)$. J. Algebra 210 (1998), no. 1, 352-360.
it seems that there is further unpulished work by Boris Weisfeiler showing that $(n+2)!$ is enough if $n \geq 64$.

EDIT: It was pointed out in a comment that the passage from finite to virtually abelian is not so straightforward.
